Epsilon-delta proof of a limit of a quadratic function I found the following example:
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*Is the reason for getting an upper bound $|x+2|\leq 3+2=5$ (and not $|x-2|\leq 5$ [I've used the triangle inqeuality here]) that we want a bound on $|x-2|$ to depend on $\epsilon$ but we don't care about a bound on $|x+2|$? So if the problem were asking to prove $\lim_ {x\to -2}=4$, I would have to say $|x-2|\leq 5$ and so $|x+2|\leq \epsilon/5$?

*If we had chosen the max $\delta$ value to be $1/2$ instead of $1$, would we have to set $\delta=\min\{1/2,\epsilon/4.5\}$?

*I'm also not sure if I fully understand the point of setting $\delta=\min\{1,\epsilon/5\}$. Let's say $\epsilon=10$. Then if I take delta to be $\epsilon/5$, I'll have $|x-2||x+2|\leq |2||5|=10$. Is the only reason why I should take $\delta=1$ in this case that the inequality $|x-2||x+2| < \epsilon$ must be strict?

 A: The key idea being used here is that "small times medium is still small". Let's show the idea not just at the point $2$, but for an arbitrary $a$.
So, we have the factorization $|x^2-a^2|=|x-a||x+a|$. The first term $|x-a|$ is nice because we can bound it by $\delta$. But what do we do with $|x+a|$? Well, the only thing we have control over is $|x-a|$, so we somehow have to "force" an $|x-a|$ term to appear from $|x+a|$. Here's what I mean:
\begin{align}
|x^2-a^2|&=|x-a|\cdot|x+a|\\
&=|x-a|\cdot |(x-a)+2a|\\
&\leq |x-a|\cdot \left(|x-a|+2|a|\right)\\
&\leq \delta\cdot (\delta+2|a|)
\end{align}
This estimate is true for all values of $a\in\Bbb{C}$, all $\delta>0$ and all $x\in\Bbb{C}$ such that $|x-a|<\delta$. Think of this as the "rough-work".
Therefore, in order to complete the proof, we have to start with a given number $\epsilon>0$, and explain why it is possible to choose $\delta>0$ such that $\delta(\delta+2|a|)<\epsilon$. Well, I can rearrange this to get $\delta<\frac{\epsilon}{\delta+2|a|}$. Ouch... I don't want my definition for $\delta$ in terms of $\epsilon$ to again depend on $\delta$, because this is very implicit, and this doesn't even tell me whether or not such a $\delta>0$  exists.
Ok, now I can try something else: $\delta(\delta+2|a|)=\delta^2+2|a|\delta=(\delta+|a|)^2-|a|^2$, so if I want this to be smaller than $\epsilon$, then by rearranging this, I find that I need $\delta<\sqrt{\epsilon+|a|^2}-|a|$. Ok, this is certainly correct, but it is undesirable for several reasons. First, how do we know that every positive number has a square root? The existence of square roots is a pretty non-trivial thing, and is likely not one of the basic things established in a course on analysis. Next, to actually deduce that $\delta>0$, we need to know that $\sqrt{\cdot}$ is actually a strictly increasing function on $[0,\infty)$. These facts might be deemed too much, because some textbooks (eg Spivak) only establishes existence of square roots after proving continuity of $x\mapsto x^2$. And the final reason why this approach is undesirable is that it's too much work; not just for the reasons I mentioned above, but also because this way, we're actually getting down to the nitty-gritty details of solving the quadratic inequality $\delta(\delta+2|a|)<\epsilon$. What if we had some more complicated expression, such as $\delta(\delta^3+\delta^2+19\delta+7)$; how can we make this $<\epsilon$?
The above approach completely ignores one of the key mantras when transitioning from calculus to analysis: we just need estimates to be "good enough". We don't need them to be "the best possible". Given an $\epsilon>0$ (for example $1000$, or $0.9$, or $6.62\times 10^{-34}$, or anything else), all we need is to find some number $\delta>0$ which satisfies the inequality $\delta(\delta+2|a|)<\epsilon$. We don't need all the possible $\delta$'s. We don't need the "best" possible $\delta$ (whatever "best" means, and in fact often there isn't even a best possible $\delta$). All we care about is finding atleast one value for $\delta$.
So, what do we do? We make a simple and crude estimate: $\delta(\delta+2|a|)\leq \delta (1+2|a|)$, provided that $\delta<1$. Next, if we want to make the second term less than $\epsilon$, then we find the condition $\delta<\frac{\epsilon}{1+2|a|}$. So, we want both conditions to be satisfied simultaneously. So, if we phrase this as a proper argument with the logic flowing in "the right direction", we can say

Given $\epsilon>0$, choose a $\delta>0$ such that $\delta<\min\left(1,\frac{\epsilon}{1+2|a|}\right)$. Then,
\begin{align}
\delta(\delta+2|a|)< \delta(1+2|a|)< \frac{\epsilon}{1+2|a|}\cdot (1+2|a|)=\epsilon.
\end{align}

We had to take $\min$ because we want a number $\delta$ satisfying three conditions simultaneously:

*

*$\delta>0$ (because that's what the definition of limits requires)

*$\delta<1$  (so the first inequality above is true)

*$\delta<\frac{\epsilon}{1+2|a|}$ (so the second inequality is true).

Note that if we have $a=2$ as in your question, then $1+2|a|=5$, so that's where the $\frac{\epsilon}{5}$ comes from.
Why did I choose the number $1$ above? Well, only because I like the number $1$, and it is a rather simple number. If I wanted, I could definitely consider $\min\left(\frac{1}{2},\frac{\epsilon}{\frac{1}{2}+2|a|}\right)$, which in the case of $a=2$ is indeed $\min\left(\frac{1}{2},\frac{\epsilon}{4.5}\right)$. This would definitely give a correct solution. Even the choice $\min\left(17,\frac{\epsilon}{17+2|a|}\right)$ would be acceptable.
Anyway, the key idea of the proof is "small times medium is small", meaning that all we have control over is $|x-a|$, which is the thing we can make "small", in other words $x$ is close to $a$. Now, if $x$ is close to $a$, then $|x+a|$ is not too large, i.e it is "medium"; and in the above proof we quantified the "mediumness" using the triangle inequality $|x+a|\leq |x-a|+2|a|\leq \delta+2|a|$, which we can make $1+2|a|$ simply by requiring $\delta<1$.
